to qualify for the indianapolis 500 race a car must must average 350 km/hr for four laps. if a driver average

Question

to qualify for the indianapolis 500 race a car must must average 350 km/hr for four laps. if a driver averages 320 km/hr for the first two laps, what average speed must be achieved for final two laps in order to qualify?

Answer 1

Approximately 386.2068966 km/hr.

If the question was to qualify for the indianapolis 500 race a car must average 350 km/hr for four minutes. if a driver averages 320 km/hr for the first two minutes, what average speed must be achieved for final two minutes in order to qualify? Then the answer would be 380 km/hr.

If you want me to show you how I solved the problem, feel free to email me.

Answer 2

enable d = the area around one lap enable t = the total time the driving force took for the 1st 9 laps. Then, on account that distance/time = velocity, the value for the 1st 9 laps is 9d/t = 195. Now enable T be the total time mandatory for all 10 laps. because of the fact the ten-lap average would desire to be 198 km/h, 10d/T = 198. The time for the final lap is T - t and so the value for the final lap is d/(T - t). From above, 9d/t = 195, or t = 9d/195. additionally, on account that 10d/T = 198, then T = 10d/198. to that end T - t = 10d/198 - 9d/195 = d(10/198 - 9/195) (T - t)/d = ((10)(195) - (9)(198)) / ((195)(198)) = (1950 - 1782)/38610 = 168/38610. hence the value mandatory for the final lap is d/(T - t) = 38610/168 = 230 km/h, to the closest km/h. Please be conscious- considered one of the different solutions potential that the value of the total ten laps is only the common of the ten speeds of all of the guy laps and promises an equation wherein the ten-lap velocity (198) is the weighed average of 9/10 of 195 and a million/10 of the final lap's velocity. it somewhat is not marvelous. the value of the final lap needless to say would desire to be bigger than 195 with a view to advance the best average to 198. yet this implies that the time for the final lap would desire to be decrease than the common of the previous lap situations. So the time for the final lap is decrease than a million/10 of the time for the total ten laps. A weighted average in accordance with variety of laps, (this is, of distance) would be too small to do the job. the marvelous answer would desire to be the weighted average in accordance with time. right here's a user-friendly occasion for instance this element: assume you have a song that's 36 km according to lap, and you % to return and forth 2 laps with a median velocity of 9 km/h. Now assume which you complete the 1st lap on the value of 6 km/h. you could think of that the 2d lap would desire to have velocity of 12 km/h on account that (6 + 12)/2 = 18/2 = 9. yet one lap at 6 km/h takes 36/6 = 6 hours. One lap at 12 km/h takes 36/12 = 3 hours. so which you have long gone 2 laps = seventy two km in 6 hr. + 3 hr. = 9 hours for a median velocity of basically seventy two/9 = 8 km/h. with a view to average 9 km/h you could desire to return and forth the 2d lap at 18 km/h. The time for the 2d lap is then 36/18 = 2 hours and th time for the two laps is 6 hr. + 2 hr. = 8 hours, and seventy two km in 8 hours ends up in a median velocity of seventy two/8 = 9 km/h.

Answer 3

380

Answer 4

380

Answer 5

only 33 cars qualify for that race all you have to do is be one of those cars..lately there anly has been about 35 or 36 cars registered for qualifying

Answer 6

not 380km/hr

Answer 7

380, thanks for the 2.